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 Post subject: Improper IntegralPosted: Fri Jul 08, 2016 1:06 pm

Joined: Sat Nov 07, 2015 6:12 pm
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Location: Larisa
Prove that: $\displaystyle \int_{-\infty }^{\infty }\frac{\cos x}{x^2+a^2}\, dx=\frac{\pi e^{-a}}{a}$ where $a>0$

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 Post subject: Re: Improper IntegralPosted: Fri Jul 08, 2016 1:07 pm

Joined: Mon Oct 26, 2015 12:27 pm
Posts: 40
Replied by ex-member aziiri:

Consider $g(z)=\frac{e^{i z}}{z^2+a^2}$, and let $C_r$ be a the half circle $\{re^{i t} | 0\leq t\leq \pi\}$ with $r>a$, then : $\int_{C_r \cup [-r,r]} g(z) \ \mathrm{d}z = 2\pi i \text{Res}(g,ia)= \frac{2\pi i e^{-a}}{2ia}= \frac{\pi e^{-a} }{a}$ And by the estimation lemma we have: $\left|\int_{C_r} g(z) \ \mathrm{d}z \right|\leq \int_{C_r} \left|\frac{e^{i z}}{z^2+a^2}\right| \ \mathrm{d}z \leq \int_{C_r} \frac{\mathrm{d}z}{r^2-a^2} = \frac{\pi r}{r^2-a^2}$ Now, take $r\to \infty$ to get : $\int\limits_{-\infty}^{+\infty}g(z) \ \mathrm{d}z = \frac{\pi e^{-a} }{a}.$ And therefore : $\int\limits_{-\infty}^{+\infty}\frac{\cos x}{x^2+a^2} \ \mathrm{d}x= \text{Re}\left(\int\limits_{-\infty}^{+\infty}g(z) \ \mathrm{d}z\right) = \frac{\pi e^{-a}}{a}.$

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 Post subject: Re: Improper IntegralPosted: Fri Jul 08, 2016 1:08 pm
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
A 2nd solution with real analysis methods:

We use that \begin{align*}
\end{align*}So
\begin{align*}
\displaystyle \int_{-\infty }^{+\infty }{\frac{\cos x}{x^2+a^2}\, dx}&=\int_{-\infty }^{0}{\frac{\cos x}{x^2+a^2}\, dx}+\int_{0}^{+\infty }{\frac{\cos x}{x^2+a^2}\, dx}\\
&\mathop{=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{u\,=\,-x}\\
{-du\,=\,dx}\\
\end{subarray}}\,\int_{0}^{+\infty}{\frac{\cos u}{u^2+a^2}\, du}+\int_{0}^{+\infty }{\frac{\cos x}{x^2+a^2}\, dx}\\
&=2\int_{0}^{+\infty }{\frac{\cos x}{x^2+a^2}\, dx}\\
&\stackrel{(1)}{=\!=}2\int_{0}^{+\infty }{\biggl({\int_{0}^{\infty }{2t\,{\mathrm{e}}^{-(x^2+a^2)t^2}\, dt}}\bigg)\cos x\, dx}\\
&=2\int_{0}^{+\infty }{2t\,{\mathrm{e}}^{-a^2t^2}\biggl({\int_{0}^{+\infty }{{\mathrm{e}}^{-x^2t^2}\cos x\,dx}}\bigg)\, dt}\\
&\stackrel{(2)}{=\!=}2\int_{0}^{+\infty }{2t\,{\mathrm{e}}^{-a^2t^2}\frac{\sqrt{\pi}}{2t}{\mathrm{e}}^{-\frac{1}{4t^2}}\, dt}\\
&=2\sqrt{\pi}\int_{0}^{+\infty}{{\mathrm{e}}^{-a^2t^2-\frac{1}{4t^2}}\, dt}\\
&\stackrel{(3)}{=\!=}2\sqrt{\pi}\,\frac{\sqrt{\pi}}{2a}\,{\mathrm{e}}^{-a}\\
&=\frac{\pi \,{\mathrm{e}}^{-a}}{a}\,.
\end{align*}

Proofs of (1), (2) $\&$ (3):

(1) Trivial.

(2) We will prove the more general $I=\displaystyle \int_{0}^{+\infty }{{\mathrm{e}}^{-t^2x^2}\cos({kx})\, dx}=\frac{\sqrt{\pi}}{2t}\,{\mathrm{e}}^{-\frac{k^2}{4t^2}}\,, \quad k\neq0\,.$ Differentiating with respect to $k$, by Leibnitz rule, we have \begin{align*}
\frac{dI}{dk}&=-\displaystyle \int_{0}^{+\infty }{x\,{\mathrm{e}}^{-t^2x^2}\sin({kx})\, dx}\\
&= \frac{1}{2t^2}\int_{0}^{+\infty }{\bigl({{\mathrm{e}}^{-t^2x^2}}\bigr)'\sin({kx})\, dx}\\
&=\biggl[{\frac{1}{2t^2}\,{\mathrm{e}}^{-t^2x^2}\sin({kx})}\biggr]_{0}^{+\infty }-\frac{k}{2t^2}\int_{0}^{+\infty }{x\,{\mathrm{e}}^{-t^2x^2}\cos({kx})\, dx}\\
I&=c_1{\mathrm{e}}^{-\frac{k^2}{4t^2}}\,.
\end{align*} For $k=0$ we have that $\displaystyle c_1{\mathrm{e}}^{-\frac{0^2}{4t^2}}=\int_{0}^{+\infty }{{\mathrm{e}}^{-t^2x^2}\cos(0\cdot x)\, dx}= \int_{0}^{+\infty }{{\mathrm{e}}^{-t^2x^2}\, dx}=\frac{\sqrt{\pi}}{2t}\,,$ and we get $c_1=\frac{\sqrt{\pi}}{2t}\,.$ So we have the desired result.

(3) Left as an exercise.

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